3.19.46 \(\int \frac {a+b x}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=69 \[ \frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \begin {gather*} \frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {a+b x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx\\ &=\frac {2}{(b d-a e) \sqrt {d+e x}}+\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{b d-a e}\\ &=\frac {2}{(b d-a e) \sqrt {d+e x}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e (b d-a e)}\\ &=\frac {2}{(b d-a e) \sqrt {d+e x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.67 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{\sqrt {d+e x} (a e-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 1, 1/2, (b*(d + e*x))/(b*d - a*e)])/((-(b*d) + a*e)*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.10, size = 79, normalized size = 1.14 \begin {gather*} \frac {2}{\sqrt {d+e x} (b d-a e)}+\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)/((d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

2/((b*d - a*e)*Sqrt[d + e*x]) + (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(-(
b*d) + a*e)^(3/2)

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fricas [A]  time = 0.46, size = 214, normalized size = 3.10 \begin {gather*} \left [-\frac {{\left (e x + d\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, \sqrt {e x + d}}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}, -\frac {2 \, {\left ({\left (e x + d\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - \sqrt {e x + d}\right )}}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

[-((e*x + d)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(
b*x + a)) - 2*sqrt(e*x + d))/(b*d^2 - a*d*e + (b*d*e - a*e^2)*x), -2*((e*x + d)*sqrt(-b/(b*d - a*e))*arctan(-(
b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - sqrt(e*x + d))/(b*d^2 - a*d*e + (b*d*e - a*e^2)
*x)]

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giac [A]  time = 0.18, size = 75, normalized size = 1.09 \begin {gather*} \frac {2 \, b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} + \frac {2}{{\left (b d - a e\right )} \sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

2*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*(b*d - a*e)) + 2/((b*d - a*e)*sqrt(x*e
+ d))

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maple [A]  time = 0.06, size = 68, normalized size = 0.99 \begin {gather*} -\frac {2 b \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}-\frac {2}{\left (a e -b d \right ) \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

-2*b/(a*e-b*d)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-2/(a*e-b*d)/(e*x+d)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 2.07, size = 57, normalized size = 0.83 \begin {gather*} -\frac {2}{\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}-\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{{\left (a\,e-b\,d\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/((d + e*x)^(3/2)*(a^2 + b^2*x^2 + 2*a*b*x)),x)

[Out]

- 2/((a*e - b*d)*(d + e*x)^(1/2)) - (2*b^(1/2)*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(a*e - b*d)^
(3/2)

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sympy [A]  time = 121.04, size = 60, normalized size = 0.87 \begin {gather*} - \frac {2}{\sqrt {d + e x} \left (a e - b d\right )} - \frac {2 \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{\sqrt {\frac {a e - b d}{b}} \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

-2/(sqrt(d + e*x)*(a*e - b*d)) - 2*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(sqrt((a*e - b*d)/b)*(a*e - b*d))

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